1
$\begingroup$

I'm studying for finals at the moment and could use some help with solving the particular solution for this system of nonhomogeneous differential equations:

$x' = \begin{bmatrix}1 & 0\\ 2 & -3 \end{bmatrix}x + \begin{bmatrix}10\cos(t)\\ 0\end{bmatrix}$

Can someone walk me through what needs to be done? I know that the particular solution will take the form: $x_p(t) = A\cos(t) + B\sin(t)$, but apart from that, I'm unsure of how to proceed.

Thanks!

$\endgroup$
  • 1
    $\begingroup$ I see you have accepted some answers, so let me help you out. Do you know how to compute matrix exponentials? It's not necessary to solve this, there is a very simple way, but computing matrix exponentials allows you to attack any kind of problem like this. $\endgroup$ – Git Gud May 1 '14 at 0:21
  • $\begingroup$ I'm not familiar with that terminology, sorry. I'm trying to go about this using the method of underdetermined coefficients. $\endgroup$ – TheColonel May 1 '14 at 0:22
  • 1
    $\begingroup$ Do you understand what's going on in this answer? $\endgroup$ – Git Gud May 1 '14 at 0:26
  • 1
    $\begingroup$ @Git Gud, I arrived at this solution for the particular solution: x_p_(t) = {{-5},{-4}}cos(t) + {{5},{2}}sin(t) Is this correct? $\endgroup$ – TheColonel May 1 '14 at 0:33
  • 1
    $\begingroup$ Your particular solution is correct. $\endgroup$ – Git Gud May 1 '14 at 0:58
1
$\begingroup$

We have:

$$\phi(t) = \left( \begin{array}{cc} 0 & 2 e^t \\ e^{-3 t} & e^t \\ \end{array} \right)$$

$$\phi^{-1}(t) = \left( \begin{array}{cc} -\frac{e^{3 t}}{2} & e^{3 t} \\ \frac{e^{-t}}{2} & 0 \\ \end{array} \right)$$

$$\phi^{-1}(t).\left( \begin{array}{c} 10 \cos (t) \\ 0 \\ \end{array} \right) = \left( \begin{array}{c} -5 e^{3 t} \cos (t) \\ 5 e^{-t} \cos (t) \\ \end{array} \right)$$

Integrating the previous result yields:

$$\left( \begin{array}{c} -\frac{1}{2} e^{3 t} (3 \cos (t)+\sin (t)) \\ \frac{5}{2} e^{-t} (\sin (t)-\cos (t)) \\ \end{array} \right)$$

So we have:

$$x_p(t) = \phi(t).\left( \begin{array}{c} -\frac{1}{2} e^{3 t} (3 \cos (t)+\sin (t)) \\ \frac{5}{2} e^{-t} (\sin (t)-\cos (t)) \\ \end{array} \right) = \left( \begin{array}{c} 5 (\sin (t)-\cos (t)) \\ 2 \sin (t)-4 \cos (t) \\ \end{array} \right)$$

So,

$$x(t) = x_h(t) + x_p(t)$$

This yields:

$$x(t) = c_1 e^t+5 \sin (t)-5 \cos (t) \\ y(t) = \frac{1}{2} \left(e^{-3 t} \left(c_1 \left(e^{4 t}-1\right)+2 c_2\right)+4 \sin (t)-8 \cos (t)\right)$$

$\endgroup$
  • $\begingroup$ Mathematica is a good thing, especially when you copy what it produces without typos. $\endgroup$ – Artem May 2 '14 at 13:39
1
$\begingroup$

Set $x=\begin{bmatrix} x_1 \\ x_2\end{bmatrix}, A=\begin{bmatrix} 1 & 0 \\ 2 & -3\end{bmatrix}$ and $b(t)=\begin{bmatrix} 10\cos (t) \\ 0\end{bmatrix}$, for all $t\in \mathbb R$.

Your system is equivalent to $x'=Ax+b$.

Given $t\in \mathbb R$, it can be rewritten as $$\begin{cases} x_1'(t)&=x_1(t)+ 10\cos (t)\\ x_2'(t)&=2x_1(t)-3x_2(t)\end{cases}$$

The equation $x_1'(t)=x_1(t)+ 10\cos (t)$ is just an ODE. It can be solved using the standard methods, (for instance this one). The general solution of this equation is determined by $$\forall t\in \mathbb R\left(x_1(t)=5\sin(t)-5\cos(t) +C_1e^t\right),$$ for some $C_1\in \mathbb R$.

So now you can just replace $x_1(t)$ by the above expression in the second equation yielding:

$$x'_2(t)=10\sin(t)-10\cos(t)+2C_1e^t-3x_2(t).$$

This, again, is just an ODE. The solution is given by, for some $C_2\in \mathbb R$, $$\forall t\in \mathbb R\left(x_2(t)=2\sin(t)-4\cos(t)+\dfrac{C_1}{2}e^t+C_2e^{-3t}\right).$$

A particular solution to the system can be found setting $C_1=C_2=0$ and this is exactly the solution you found.


Since you're looking to solve this with the indeterminate coefficients method, let me also do this.

Take the ansatz $x_p=B\cos +C\sin $, where $B,C\in \mathbb R^{2\times 1}$ are constants.
Write $B=\begin{bmatrix} b_1\\ b_2\end{bmatrix}, C=\begin{bmatrix} c_1\\ c_2\end{bmatrix}$ and $x_p(t)=\begin{bmatrix} b_1\cos(t)+c_1\sin (t)\\ b_2\cos(t)+c_2\sin(t)\end{bmatrix}$

For all $t\in \mathbb R$ set $$\begin{bmatrix} -b_1\sin(t)+c_1\cos (t)\\ -b_2\sin(t)+c_2\cos(t)\end{bmatrix}=x_p'(t)=Ax_p(t)+b(t)=\begin{bmatrix} b_1\cos(t) + c_1\sin(t)+10\cos(t)\\ (2b_1-3b_2)\cos(t)+(2c_1-3c_2)\sin(t)\end{bmatrix},$$

which is equivalent to $$\begin{cases} (10+b_1-c_1)\cos(t)+(c_1+b_1)\sin(t)&=0\\ (2b_1-3b_2-c_2)\cos(t)+(2c_1-3c_2+b_2)\sin(t)&=0\end{cases} \tag {$\spadesuit$}$$

Since $\left(\spadesuit\right)$ holds for all $t$, it holds in particular for $t=0$ and $t=\dfrac\pi 2$ thus providing the system $$\begin{cases} b_1+ c_1&=0\\ b_1-c_1&=-10\\ 2b_1-3b_2-c_2&=0\\ b_2+2c_1-3c_2&=0\end{cases}$$

whose solution is attained with $(b_1, b_2, c_1, c_2)=(-5, -4, 5, 2)$.

Therefore $x_p(t)=\begin{bmatrix} -5\cos(t)+5\sin (t)\\ -4\cos(t)+2\sin(t)\end{bmatrix},$ for all $t\in \mathbb R$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.